Question

A cylinder initially contains 1.5 kg of air at 100 kPa and 17 C. The air is compressed polytropically until the volume is reduced by one-half. The polytropic exponent is 1.3. Determine the work done (absolute value) and heat transfer for this process.

Answer 1

Answer:

Work done = 96.15 KJ

Heat transferred = 24 KJ

Explanation:

We are given;

Mass of air;m = 1.5 kg

Initial pressure;P1 = 100 KPa

Initial temperature;T1 = 17°C = 273 + 17 K = 290 K

Final volume;V2 = 0.5V1

Since the polytropic exponent is 1.3,thus it means;

P2 = P1[V1/V2]^(1.3)

So,P2 = 100(V1/0.5V1)^(1.3)

P2 = 100(2)^(1.3)

P2 = 246.2 KPa

To find the final temperature, we will make use of combined gas law.

So,

(P1×V1)/T1 = (P2×V2)/T2

T2 = (P2×V2×T1)/(P1×V1)

Plugging in the known values;

T2 = (246.2×0.5V1×290)/(100×V1)

V1 cancels out to give;

T2 = (246.2×0.5×290)/100

T2 = 356.99 K

The boundary work for this polytropic process is given by;

W_b = - ∫P. dv between the initial boundary and final boundary.

Thus,

W_b = - (P2.V2 - P1.V1)/(1 - n) = -mR(T2 - T1)/(1 - n)

R is gas constant for air = 0.28705 KPa.m³/Kg.K

n is the polytropic exponent which is 1.3

Thus;

W_b = -1.5 × 0.28705(356.99 - 290)/(1 - 1.3)

W_b = 96.15 KJ

The formula for the heat transfer is given as;

Q_out = W_b - m.c_v(T2 - T1)

c_v for air = 0.718 KJ/Kg.k

Q_out = 96.15 - (1.5×0.718(356.99 - 290))

Q_out = 24 KJ